HDU 5919

HDU 5919

题意：

　　动态处理一个序列的区间问题，对于一个给定序列，每次输入区间的左端点和右端点，输出这个区间中：每个数字第一次出现的位子留下， 输出这些位子中最中间的那个，就是（len+1）/2那个。

思路：

　　主席树操作，这里的思路是从n到1开始建树。其他就是主席树查询区间第K小，计算区间不同值个数。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
#include <unordered_map>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue
#define max3(a,b,c) max(max(a,b),c)



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;       
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 256;
const double esp = 1e-8;
const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/   
            const int maxn = 2e5+9;
            struct node {
                int l,r;
                int sum;
            } t[maxn*40];
            int tot;
            int vis[maxn],rt[maxn],a[maxn];

            void update(int l,int r,int &x, int y,int pos,int val){
                tot++; x = tot;
                t[x] = t[y];
                t[x].sum += val;
                if(l == r)return;
                int mid = (l + r) >> 1;
                if(pos <= mid) update(l,mid,t[x].l,t[y].l,pos,val);
                else update(mid+1,r,t[x].r,t[y].r,pos,val);
            }

            int query(int l,int r,int x,int L, int R){
                if(l >= L && r<=R){
                    return t[x].sum;
                }
                int mid = (l + r) >> 1;
                int res = 0;
                if(mid >= L) res += query(l,mid,t[x].l,L,R);
                if(mid <  R) res += query(mid+1,r,t[x].r,L,R);
                return res;
            }

            int getk(int l,int r,int x,int k){
                if(l == r)return l;
                int cnt = t[t[x].l].sum;
                int mid = (l+r)>>1;
                if(cnt >= k) return getk(l,mid,t[x].l,k);
                else return getk(mid+1, r,t[x].r,k-cnt);
            }
int main(){
            int T,n,m;  scanf("%d", &T);
            for(int tt=1; tt<=T; tt++){
                tot = 0;
                scanf("%d%d", &n, &m);
                for(int i=1; i<=n; i++)scanf("%d", &a[i]);
                memset(vis, -1, sizeof(vis));
                memset(rt, 0, sizeof(rt));
                for(int i=n; i>=1; i--){
                    if(vis[a[i]] == -1){
                        update(1,n,rt[i],rt[i+1],i,1);
                    }
                    else{
                        int tmp;
                        update(1,n,tmp,rt[i+1], vis[a[i]], -1);
                        update(1,n,rt[i],tmp,i,1);
                    }
                    vis[a[i]] = i;
                }
                printf("Case #%d:", tt);
                int la = 0;
                for(int i=1; i<=m; i++){
                    int l,r,L,R;    scanf("%d%d", &L, &R);
                    l = min((L+la)%n + 1, (R+la)%n + 1);
                    r = max((L+la)%n + 1, (R+la)%n + 1);
                    int k = query(1,n,rt[l],l,r);
                    k = (k + 1) / 2;
                    la = getk(1,n,rt[l],k);
                    printf(" %d", la);
                }
                puts("");
            }
            return 0;
}